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            052184472Xc09  CUNY148/Severini  June 2, 2005  12:8





                                                  9.3 Asymptotic Expansions                  267

                                            Table 9.2. Approximations in Example 9.2.

                                                                       Relative
                                          x       Exact     Approx.    error (%)

                                          0.50    0.102     0.103      1.4
                                          0.20    0.0186    0.0186     0.12
                                          0.10    0.00483   0.00483    0.024
                                          0.05    0.00123   0.00123    0.0056




                          Table 9.2 contains values of the approximation
                                              1        β−1     β − 1  x
                                                α
                                               x (1 − x)   1 +
                                              α                α + 1 1 − x
                        for the case α = 2, β = 3/2, along with the exact value of the integral and the relative
                        error of the approximation. Note that for these choices of α and β,exact calculation of the
                        integral is possible:
                                           x
                                                 1      4   2       5  2      3
                                           t(1 − t) 2 dt =  + (1 − x) 2 − (1 − x) 2 .
                                         0             15   5          3
                        The results in Table 9.2 show that the approximation is extremely accurate even for relatively
                        large values of x.



                        Example 9.3 (Normal tail probability). Consider the function
                                                            ∞

                                                    ¯
                                                     (z) ≡    φ(t) dt
                                                           z
                        where φ denotes the density function of the standard normal distribution. Then, using
                        integration-by-parts, we may write
                                                   ∞           ∞  1

                                           ¯
                                            (z) =    φ(t) dt =     tφ(t) dt
                                                  z           z   t
                                                   1      ∞     ∞  1

                                                           −       φ(t) dt
                                                   t     z    z  t
                                               =− φ(t)            2
                                                  1         ∞  1
                                               =   φ(z) −      tφ(t) dt
                                                  z       z  t  3
                                                  1      1          ∞  1
                                               =   φ(z) −  φ(z) +      φ(t) dt.
                                                  z      z 3      z  t 4
                        Hence,

                                                  1       1          1
                                            ¯                            ¯
                                            (z) =  φ(z) −  φ(z) + O      (z)
                                                  z       z 3        z 4
                        as z →∞. That is,

                                                    1               1   1
                                                         ¯
                                             1 + O       (z) = φ(z)   −    ,
                                                    z 4             z   z 3
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